Answer: 4 s
Explanation:
Given
The ball leaves the hand of student with a speed of 
When the hand is
above the ground
Using the equation of motion we can write

Substitute the values
![\Rightarrow 2.5=-19t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-19t-2.5=0\\\\\Rightarrow t=\dfrac{19\pm \sqrt{(-19)^2-4\times 4.9\times (-2.5)}}{2\times 19}\\\Rightarrow t=4.0049\quad [\text{Neglecting the negative value of }t]](https://tex.z-dn.net/?f=%5CRightarrow%202.5%3D-19t%2B0.5%5Ctimes%209.8t%5E2%5C%5C%5CRightarrow%204.9t%5E2-19t-2.5%3D0%5C%5C%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B19%5Cpm%20%5Csqrt%7B%28-19%29%5E2-4%5Ctimes%204.9%5Ctimes%20%28-2.5%29%7D%7D%7B2%5Ctimes%2019%7D%5C%5C%5CRightarrow%20t%3D4.0049%5Cquad%20%5B%5Ctext%7BNeglecting%20the%20negative%20value%20of%20%7Dt%5D)
Thus, the ball will take 4 s to hit the ground.
r1 = 5*10^10 m , r2 = 6*10^12 m
v1 = 9*10^4 m/s
From conservation of energy
K1 +U1 = K2 +U2
0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2
0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2
M is mass of sun = 1.98*10^30 kg
G = 6.67*10^-11 N.m^2/kg^2
0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))
v2 = 5.35*10^4 m/s
Answer:
The inquiry process takes advantage of the natural human desire to make sense of the world... This attitude of curiosity permeates the inquiry process and is the fuel that allows it to continue. Process skills are not used for their own sake.
A. Forced vibrations, such as those between a tuning fork and a large cabinet surface, result in a much lower sound than was produced by the original vibrating body.
Answer:
a)30.14 rad/s2
b)43.5 rad/s
c)60633 J
d)42 kW
e)84 kW
Explanation:
If we treat the propeller is a slender rod, then its moments of inertia is

a. The angular acceleration is Torque divided by moments of inertia:

b. 5 revolution would be equals to
rad, or 31.4 rad. Since the engine just got started


c. Work done during the first 5 revolution would be torque times angular displacement:

d. The time it takes to spin the first 5 revolutions is

The average power output is work per unit time
or 42 kW
e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:
or 84 kW